Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $q = \dfrac{-4a^2 + 28a}{5a^3 + 75a^2 + 280a} \div \dfrac{-3a + 27}{a^2 - 2a - 63} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-4a^2 + 28a}{5a^3 + 75a^2 + 280a} \times \dfrac{a^2 - 2a - 63}{-3a + 27} $ First factor out any common factors. $q = \dfrac{-4a(a - 7)}{5a(a^2 + 15a + 56)} \times \dfrac{a^2 - 2a - 63}{-3(a - 9)} $ Then factor the quadratic expressions. $q = \dfrac {-4a(a - 7)} {5a(a + 7)(a + 8)} \times \dfrac {(a + 7)(a - 9)} {-3(a - 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {-4a(a - 7) \times (a + 7)(a - 9) } { 5a(a + 7)(a + 8) \times -3(a - 9)} $ $q = \dfrac {-4a(a + 7)(a - 9)(a - 7)} {-15a(a + 7)(a + 8)(a - 9)} $ Notice that $(a + 7)$ and $(a - 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {-4a\cancel{(a + 7)}(a - 9)(a - 7)} {-15a\cancel{(a + 7)}(a + 8)(a - 9)} $ We are dividing by $a + 7$ , so $a + 7 \neq 0$ Therefore, $a \neq -7$ $q = \dfrac {-4a\cancel{(a + 7)}\cancel{(a - 9)}(a - 7)} {-15a\cancel{(a + 7)}(a + 8)\cancel{(a - 9)}} $ We are dividing by $a - 9$ , so $a - 9 \neq 0$ Therefore, $a \neq 9$ $q = \dfrac {-4a(a - 7)} {-15a(a + 8)} $ $ q = \dfrac{4(a - 7)}{15(a + 8)}; a \neq -7; a \neq 9 $